Physics Alternating Current

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New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

X c = 1 ω c , X L = ω L so at very high frequencies capacitor behaves as conduct and inductor behaves as open circuit. The effective impedance will be 2 Ω .

 

New answer posted

5 months ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

Z = R 2 + ( X C X L ) 2

Z = R 2 + ( R R ) 2 [ ? X L = X C = R G i v e n ]

Z = R

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

c o s ? 1 = R R 2 + ( 3 R ) 2 = 1 1 0

c o s ? 2 = R R 2 + ( 3 R R ) 2 = 1 2

c o s ? 2 c o s ? 1 = 1 0 2 = 5 x = 1

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

f r m s 2 = ( 4 2 2 ) 2 + 1 0 2 = 1 2 1

f r m s = 1 1 A

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

Δ V D 1 = Δ V D 2 = l * R = 0

Since resistance is zero in forward bias.

=> V0 = 5V

New answer posted

5 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Let, current through inductor,

i L = l L s i n ( ω t + ? 1 )                

l L = 2 0 0 2 5 0 2 + ( 1 0 0 * 0 . 5 0 ) 2 = 4 A              

? 1 = t a n 1 ( 1 0 0 * 0 . 5 0 5 0 ) = π 4             

Let, current through capacitor,

As  ? 2 ? 1 = π 2  

l = ( I L ) 2 + ( I C ) 2

= 4 2 + 2 2 = 4 . 4 7 A              

             

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Power is maximum at resonance

So, X L = X C = 1 ω C

C = 1 ω * X L

= 1 2 5 0 * 1 0 0 * π 2

= 4 μ F [ π 2 = 1 0 ]

New answer posted

5 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Purely inductive circuit

θ = π 2

c o s π 2 = 0

Average power = 0

New answer posted

5 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

ω = 2 π f = 2 π * 5 0 = 1 0 0 π

X L = w L = 1 0 0 π * 1 0 0 * 1 0 3

X C = 1 w C = 1 1 0 0 π * 1 0 0 * 1 0 6 = 1 0 0 π

z = 1 0 . 0 0 8 6 Ω

l = v z = 2 2 0 1 0 . 0 0 8 = 2 2 A

New answer posted

5 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

V = 210 sin 3000 t

ω = 3000

X L = ω L = 3 0                

X C = 1 ω c = 4 0 3                

t a n ? = V L V C V R = X L X C R                

t a n ? = 0 . 1 7 ? = t a n 1 ( 0 . 1 7 )                

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